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Chapter 22: Problem 66

Draw the isomer(s) specified. There may be more than one possible isomer for each part. a. a cyclic compound that is an isomer of trans-2-butene b. an ester that is an isomer of propanoic acid c. a ketone that is an isomer of butanal d. a secondary amine that is an isomer of butylamine e. a tertiary amine that is an isomer of butylamine f. an ether that is an isomer of 2-methyl-2-propanol g. a secondary alcohol that is an isomer of 2-methyl-2-propanol

Short Answer

Expert verified
a. Cyclobutane: -C- / \ C C \ / -C- b. Methyl ethanoate: H3C - O - C(=O) - CH3 c. Propanone: H3C - C(=O) - CH3 d. N-methylpropylamine: H3C - CH2 - CH2 - NH - CH3 e. N,N-dimethylethylamine: H3C - N(CH3)2 - CH2 - CH3 f. Methoxypropane: H3C - O - CH2 - CH3 - CH3 g. 2-butanol: H3C - CH2 - CH(OH) - CH3

Step by step solution

01

a. Cyclic compound isomer of trans-2-butene

First, find the molecular formula of trans-2-butene. It is an alkene, with the formula C4H8. Our goal is to draw a cyclic isomer of this compound. A cyclic isomer should be an alkane with the same formula C4H8. We can draw cyclobutane: -C- / \ C C \ / -C- Cyclobutane is a cyclic compound that is an isomer of trans-2-butene.
02

b. Ester isomer of propanoic acid

Propanoic acid is a carboxylic acid with the formula C3H6O2. To find an ester isomer, rewrite the molecular formula in the ester format, which is RCOOR'. We have a methyl ester (C2H5COOCH3). The ester isomer of propanoic acid is methyl ethanoate: H3C - O - C(=O) - CH3
03

c. Ketone isomer of butanal

Butanal, an aldehyde, has the molecular formula C4H8O. A ketone isomer should also have the same molecular formula, C4H8O. We have propanone. The ketone isomer of butanal is propanone: H3C - C(=O) - CH3
04

d. Secondary amine isomer of butylamine

Butylamine has the formula C4H9NH2, which is a primary amine. To find a secondary amine isomer, we will have the same molecular formula C4H10N. The secondary amine isomer can be N-methylpropylamine. The secondary amine isomer of butylamine is N-methylpropylamine: H3C - CH2 - CH2 - NH - CH3
05

e. Tertiary amine isomer of butylamine

Now, we have to find a tertiary amine isomer with the same molecular formula C4H10N. The tertiary amine isomer can be N,N-dimethylethylamine. The tertiary amine isomer of butylamine is N,N-dimethylethylamine: H3C - N(CH3)2 - CH2 - CH3
06

f. Ether isomer of 2-methyl-2-propanol

2-methyl-2-propanol, a tertiary alcohol, has the molecular formula C4H10O. An ether isomer should also have this molecular formula C4H10O. We have methoxypropane. The ether isomer of 2-methyl-2-propanol is methoxypropane: H3C - O - CH2 - CH3 - CH3
07

g. Secondary alcohol isomer of 2-methyl-2-propanol

We already know that 2-methyl-2-propanol has the molecular formula C4H10O. A secondary alcohol isomer should also have this molecular formula C4H10O. We have 2-butanol. The secondary alcohol isomer of 2-methyl-2-propanol is 2-butanol: H3C - CH2 - CH(OH) - CH3

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Most popular questions from this chapter

Glucose can occur in three forms: two cyclic forms and one open-chain structure. In aqueous solution, only a tiny fraction of the glucose is in the open-chain form. Yet tests for the presence of glucose depend on reaction with the aldehyde group, which is found only in the open-chain form. Explain why these tests work.

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