Solution Found!

Compound A has molecular formula C7H14O and reacts with

Chapter 20, Problem 20.76

(choose chapter or problem)

Get Unlimited Answers! Check out our subscriptions
QUESTION:

Compound A has molecular formula C7H14O and reacts with sodium borohydride in methanol to form an alcohol. The 1 H NMR spectrum of compound A exhibits only two signals: a doublet (I = 12) and a septet (I = 2). Treating compound A with 1,2-ethanedithiol (HSCH2CH2SH) under acidic conditions, followed by Raney nickel gives compound B. (a) How many signals will appear in the 1 H NMR spectrum of compound B? (b) How many signals will appear in the 13C NMR spectrum of compound B? (c) Describe how you could use IR spectroscopy to verify the conversion of compound A to compound B.

Not The Solution You Need? Search for Your Answer Here:

Questions & Answers

QUESTION:

Compound A has molecular formula C7H14O and reacts with sodium borohydride in methanol to form an alcohol. The 1 H NMR spectrum of compound A exhibits only two signals: a doublet (I = 12) and a septet (I = 2). Treating compound A with 1,2-ethanedithiol (HSCH2CH2SH) under acidic conditions, followed by Raney nickel gives compound B. (a) How many signals will appear in the 1 H NMR spectrum of compound B? (b) How many signals will appear in the 13C NMR spectrum of compound B? (c) Describe how you could use IR spectroscopy to verify the conversion of compound A to compound B.

ANSWER:

Step 1 of 5

First, note that the molecular formula of compound A indicates a hydrogen deficiency index of one. So, one ring or one double bond should be present in the molecule. The fact that compound A reacts with sodium borohydride in methanol to form an alcohol suggests that it’s an aldehyde or ketone; this carbonyl group would explain the element of unsaturation.

The presence of both a doublet and a septet (integrations of 12 and 2, respectively) is very characteristic of isopropyl groups. The two isopropyl groups must be symmetrical, as their signals appear in the same region of the spectrum. Since each isopropyl group requires three carbons and seven hydrogens, we’re left with only one carbon and one hydrogen atom in the molecule. This indicates a ketone is present, and not an aldehyde. Hence, the structure of compound A is shown as follows:

                                                 

 

Add to cart


Study Tools You Might Need

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back