(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)-W/WILEYPLUS
- For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. དྲ。 ✗MgBr ? O CI Will the first product that forms in this reaction create a new C-C bond? Yes No • ? Will the first product that forms in this reaction create a new CC bond? Yes No × : ☐ Xarrow_forwardPredict the major products of this organic reaction: OH NaBH4 H ? CH3OH Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. ☐ : Sarrow_forwardPredict the major products of this organic reaction: 1. LIAIHA 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. X : ☐arrow_forward
- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
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- 3-Oxo-butanenitrile and (E)-2-butenal are mixed with sodium ethoxide in ethanol. Draw and name the structures of the products.arrow_forwardWhat is the reason of the following(use equations if possible) a.) In MO preperation through diazotization: Addition of sodium nitrite in acidfied solution in order to form diazonium salt b.) in MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at low pH c.) In MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at pH 4.5 d.) Avoiding not cooling down the reaction mixture when preparing the diazonium salt e.) Cbvcarrow_forwardA 0.552-g sample of an unknown acid was dissolved in water to a total volume of 20.0 mL. This sample was titrated with 0.1103 M KOH. The equivalence point occurred at 29.42 mL base added. The pH of the solution at 10.0 mL base added was 3.72. Determine the molar mass of the acid. Determine the Ka of the acid.arrow_forward
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