(a)
Interpretation: For a given set of nitrogen containing compounds, the structures have to be drawn using their general or IUPAC names.
Concept Introduction: If ,
and
groups are attached to the parent carbon, they are called primary, secondary and tertiary
There are two ways followed to name the compound. First one is the method of giving general name in which name of the alkyl group followed by amine name is given. Second one is the method giving IUPAC name in which name of the alkane group followed by amine name is given.
The length of the chain which is having more number of carbon atoms is selected as the parent or main chain. Other chains are considered as substituents to the main chain. Position of the substituents should be included in the name. If one, two, three, four, five, six, etc carbons are activating as the main chain in IUPAC system, then the name of the compound comes as methane, ethane, propane, butane, pentane, hexane, etc. which are the name of
If any configuration is present in the compound, that should be assigned to it while writing the name. If the compound contains heavier groups on the same side, it gets (Z)-configuration. If they are on the opposite directions, (E)-configurations result.
If a carbon has four different groups attached to it, that carbon shows a chirality nature. If that chiral carbon rotates the plane polarized light into a clockwise direction, it gets (R)-isomer. If that carbon rotates the plane polarized light into a counter-clockwise direction, it gets (S)-isomer.
To find: Categorize the number of alkyl groups attached to nitrogen atom
(b)
Interpretation: For a given set of nitrogen containing compounds, the structures have to be drawn using their general or IUPAC names.
Concept Introduction: If ,
and
groups are attached to the parent carbon, they are called primary, secondary and tertiary amines respectively.
There are two ways followed to name the compound. First one is the method of giving general name in which name of the alkyl group followed by amine name is given. Second one is the method giving IUPAC name in which name of the alkane group followed by amine name is given.
The length of the chain which is having more number of carbon atoms is selected as the parent or main chain. Other chains are considered as substituents to the main chain. Position of the substituents should be included in the name. If one, two, three, four, five, six, etc carbons are activating as the main chain in IUPAC system, then the name of the compound comes as methane, ethane, propane, butane, pentane, hexane, etc. which are the name of alkanes. If one, two, three, four, five, six, etc carbons are activating as the main chain in general name method, then the name of the compound comes as methyl, ethyl, propyl, butyl, pentyl, hexyl, etc. which are the name of alkyl groups. If substituent groups are attached to nitrogen atom as in the case of tertiary or secondary amines, the name is given as N-alkyl name of the substituent.
If any configuration is present in the compound, that should be assigned to it while writing the name. If the compound contains heavier groups on the same side, it gets (Z)-configuration. If they are on the opposite directions, (E)-configurations result.
If a carbon has four different groups attached to it, that carbon shows a chirality nature. If that chiral carbon rotates the plane polarized light into a clockwise direction, it gets (R)-isomer. If that carbon rotates the plane polarized light into a counter-clockwise direction, it gets (S)-isomer.
To find: Categorize the number of alkyl groups attached to nitrogen atom
In the given compound (b), three alkyl groups are attached to nitrogen atom. Therefore, it belongs to tertiary amine.
Decide the given name which is in general name method or IUPAC name method
There are three alkyl groups present in the given name. Here, the main chain is cyclopropylamine. Three-membered carbon ring structure is called cyclopropyl group. So, general name system is followed.
Locate the substituents and draw the corresponding structure for the given name
(c)
Interpretation: For a given set of nitrogen containing compounds, the structures have to be drawn using their general or IUPAC names.
Concept Introduction: If ,
and
groups are attached to the parent carbon, they are called primary, secondary and tertiary amines respectively.
There are two ways followed to name the compound. First one is the method of giving general name in which name of the alkyl group followed by amine name is given. Second one is the method giving IUPAC name in which name of the alkane group followed by amine name is given.
The length of the chain which is having more number of carbon atoms is selected as the parent or main chain. Other chains are considered as substituents to the main chain. Position of the substituents should be included in the name. If one, two, three, four, five, six, etc carbons are activating as the main chain in IUPAC system, then the name of the compound comes as methane, ethane, propane, butane, pentane, hexane, etc. which are the name of alkanes. If one, two, three, four, five, six, etc carbons are activating as the main chain in general name method, then the name of the compound comes as methyl, ethyl, propyl, butyl, pentyl, hexyl, etc. which are the name of alkyl groups. If substituent groups are attached to nitrogen atom as in the case of tertiary or secondary amines, the name is given as N-alkyl name of the substituent.
If any configuration is present in the compound, that should be assigned to it while writing the name. If the compound contains heavier groups on the same side, it gets (Z)-configuration. If they are on the opposite directions, (E)-configurations result.
If a carbon has four different groups attached to it, that carbon shows a chirality nature. If that chiral carbon rotates the plane polarized light into a clockwise direction, it gets (R)-isomer. If that carbon rotates the plane polarized light into a counter-clockwise direction, it gets (S)-isomer.
To find: Categorize the number of alkyl groups attached to nitrogen atom
In the given compound (c), only one alkyl group, namely pentane is attached to nitrogen atom. Therefore, it belongs to primary amine.
Decide the given name which is in general name method or IUPAC name method
There is only one alkane group present in the given name. It is pentane attached to primary amine named as pentanamine which is the main chain of the given name. So, IUPAC name system is followed.
Locate the substituents and draw the corresponding structure for the given name
(d)
Interpretation: For a given set of nitrogen containing compounds, the structures have to be drawn using their general or IUPAC names.
Concept Introduction: If ,
and
groups are attached to the parent carbon, they are called primary, secondary and tertiary amines respectively.
There are two ways followed to name the compound. First one is the method of giving general name in which name of the alkyl group followed by amine name is given. Second one is the method giving IUPAC name in which name of the alkane group followed by amine name is given.
The length of the chain which is having more number of carbon atoms is selected as the parent or main chain. Other chains are considered as substituents to the main chain. Position of the substituents should be included in the name. If one, two, three, four, five, six, etc carbons are activating as the main chain in IUPAC system, then the name of the compound comes as methane, ethane, propane, butane, pentane, hexane, etc. which are the name of alkanes. If one, two, three, four, five, six, etc carbons are activating as the main chain in general name method, then the name of the compound comes as methyl, ethyl, propyl, butyl, pentyl, hexyl, etc. which are the name of alkyl groups. If substituent groups are attached to nitrogen atom as in the case of tertiary or secondary amines, the name is given as N-alkyl name of the substituent.
If any configuration is present in the compound, that should be assigned to it while writing the name. If the compound contains heavier groups on the same side, it gets (Z)-configuration. If they are on the opposite directions, (E)-configurations result.
If a carbon has four different groups attached to it, that carbon shows a chirality nature. If that chiral carbon rotates the plane polarized light into a clockwise direction, it gets (R)-isomer. If that carbon rotates the plane polarized light into a counter-clockwise direction, it gets (S)-isomer.
To find: Categorize the number of alkyl groups attached to nitrogen atom
In the given compound (d), only one alkyl group is attached to nitrogen atom. Therefore, it belongs to primary amine.
Decide the given name which is in general name method or IUPAC name method
There is only one alkyl group present in the given name. It is benzyl group. So, simple general name system is followed. If phenyl group is attached to methylene group, it is called benzyl group.
Locate the substituents and draw the corresponding structure for the given name

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Chapter 22 Solutions
ORGANIC CHEMISTRY (LL) >CUSTOM PACKAGE<
- Predict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. D ㄖˋ ید H No reaction. + 5 H₂O.* Click and drag to start drawing a structure. OH H₂Oarrow_forwardDraw one product of an elimination reaction between the molecules below. Note: There may be several correct answers. You only need to draw one of them. You do not need to draw any of the side products of the reaction 'O 10 + x 也 HO + 义 Click and drag to start drawing a structure.arrow_forwardWhat are the angles a and b in the actual molecule of which this is a Lewis structure? H- :0: C=N: b Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes. a = 0° b=0 Xarrow_forward
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more products missing from the right-hand side, but there are no reagents missing from the left-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing products to the right-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + This transformation can't be done in one step. T iarrow_forwardDetermine the structures of the missing organic molecules in the following reaction: H+ O OH H+ + H₂O ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structure of the missing organic molecule X. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. X § ©arrow_forwardTable 1.1 Stock Standard Solutions Preparation. The amounts shown should be dissolved in 100 mL. Millipore water. Calculate the corresponding anion concentrations based on the actual weights of the reagents. Anion Amount of reagent (g) Anion Concentration (mg/L) 0.1649 Reagent Chloride NaCl Fluoride NaF 0.2210 Bromide NaBr 0.1288 Nitrate NaNO3 0.1371 Nitrite NaNO2 0.1500 Phosphate KH2PO4 0.1433 Sulfate K2SO4 0.1814arrow_forward
- Draw the structure of the pound in the provided CO as a 300-1200 37(2), 11 ( 110, and 2.5 (20arrow_forwardPlease help me with # 4 and 5. Thanks in advance!arrow_forwardA small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…arrow_forward
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:arrow_forward3:48 PM Fri Apr 4 K Problem 4 of 10 Submit Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Mg. :0: Select to Add Arrows :0: :Br: Mg :0: :0: Select to Add Arrows Mg. Br: :0: 0:0- Br -190 H 0:0 Select to Add Arrows Select to Add Arrows neutralizing workup H CH3arrow_forwardIarrow_forward
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