
Concept explainers
(a)
Interpretation:
The type of isomerism that could be exhibited by the given formulae and an example that illustrates the specific type of isomerism has to be stated.
Concept introduction:
Organic compounds which have similar chemical formula but different structures, i.e., the atoms are in a different spatial arrangement, they are known as structural isomers. Cis and Trans isomers are also structural or geometrical isomers.
(a)

Answer to Problem 2Q
Explanation of Solution
The type of isomerism that could be exhibited by
The structural isomer of
The given formula,
The root word hexene indicates that six carbon atoms are present in the parent chain. Double bond is present between first and second carbon.
The structure of 1-hexene is shown
The structural isomer of
The given formula,
The root word hexene indicates that six carbon atoms are present in the parent chain. The number
The structure of 2-hexene is shown
The structural isomer of
The given formula,
The root word hexene indicates that six carbon atoms are present in the parent chain. The number
The structure of 3-hexene is shown
The structural isomer of
The given formula,
The root word pentene indicates that five carbon atoms are present in the parent chain. The number
The structure of
The structural isomer of
The given formula,
The root word butene indicates that six carbon atoms are present in the parent chain. The double bond is present between second and third carbon. Two methyl groups are attached at second and third carbon.
The structure of
The geometric isomer of
The structure of cis-2-hexene is,
The structure of trans-2-hexene is,
In the cis isomer, the two hydrogen of the double bonded carbons are adjacent to each other while in trans isomer, they are opposite to each other.
The geometric isomer of
The structure of cis-3-hexene is,
The structure of trans-3-hexene is,
In the cis isomer, the two hydrogen of the double bonded carbons are adjacent to each other while in trans isomer, they are opposite to each other.
(b)
Interpretation:
The type of isomerism that could be exhibited by the given formulae and an example that illustrates the specific type of isomerism has to be stated.
Concept introduction:
Organic compounds which have similar chemical formula but different structures, i.e., the atoms are in a different spatial arrangement, they are known as structural isomers. Cis and Trans isomers are also structural or geometrical isomers.
(b)

Answer to Problem 2Q
Explanation of Solution
The type of isomerism that could be exhibited by
The structural isomer of
The root word pentane indicates that five carbon atoms are present in the parent chain. Hydroxyl group is attached to first carbon.
The structure of pentan-1-ol is shown,
The structural isomer of
The root word butane indicates that four carbon atoms are present in the parent chain.
The structure of
The structural isomer of
The root word propane indicates that three carbon atoms are present in the parent chain.
The structure of 2-ethoxypropane is shown
The structural isomer of
The root word butane indicates that four carbon atoms are present in the parent chain.
The structure of 2-methoxybutane is shown
The structural isomer of
The root word pentane indicates that five carbon atoms are present in the parent chain. Hydroxyl group is attached to third carbon.
The structure of pentan-
The structural isomer of
The root word pentane indicates that five carbon atoms are present in the parent chain. Hydroxyl group is attached to second carbon.
The structure of pentan-
(c)
Interpretation:
The type of isomerism that could be exhibited by the given formulae and an example that illustrates the specific type of isomerism has to be stated.
Concept introduction:
Organic compounds which have similar chemical formula but different structures, i.e., the atoms are in a different spatial arrangement, they are known as structural isomers. Cis and Trans isomers are also structural or geometrical isomers.
(c)

Answer to Problem 2Q
Explanation of Solution
The type of isomerism that could be exhibited by
The structural isomer of
In
The structural isomer of
In
The structural isomer of
In
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Chapter 22 Solutions
Chemistry
- Identify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forwardPlease help with number 6 I got a negative number could that be right?arrow_forward1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forward
- Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forwardFive isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forward
- Draw the Zaitsev product famed when 2,3-dimethylpentan-3-of undergoes an El dehydration. CH₂ E1 OH H₁PO₁ Select Draw Templates More QQQ +H₂Oarrow_forwardComplete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forward
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-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 reactants to the left-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. + T G OH де OH This transformation can't be done in one step.arrow_forwardMacmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forwardDraw the condensed structure of 3-hydroxy-2-butanone. Click anywhere to draw the first atom of your structure.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning


