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Topic: Scheme (a shorter one)  (Read 3706 times)

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Offline Rutherford

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Scheme (a shorter one)
« on: October 24, 2012, 09:29:18 AM »
I need some help here. Got that:
a=HNO3+H2SO4
b=Fe+H+
c=NaNO2+HCl
d= I thought benzoic acid, so an ester is made, but in the answer it is benzoylchloride. Why?
I don't understand the reaction with Br2/CH3COOH. Why is acetic acid added and how do I know where the bromine will be substituted? Could it as an electrophile attack the carbonyl group?
That's for now.
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Offline synthnick

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Re: Scheme (a shorter one)
« Reply #1 on: October 24, 2012, 10:03:43 AM »
Br2/CH3COOH is the standard conditions for α-bromination of ketones.
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Offline discodermolide

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Re: Scheme (a shorter one)
« Reply #2 on: October 24, 2012, 10:13:38 AM »
Only when there is an alpha hydrogen present.
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Offline Rutherford

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Re: Scheme (a shorter one)
« Reply #3 on: October 24, 2012, 12:49:04 PM »
Thanks, didn't know that. Then B has to be C6H5COO-C6H4-COCH2Cl
The only two things that bother me now are d and the conversion from B to C. Could someone explain these?
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Offline discodermolide

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Re: Scheme (a shorter one)
« Reply #4 on: October 24, 2012, 01:02:09 PM »
d looks like an esterification with benzoyl chloride.
I'll think about the other one.
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Offline Rutherford

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Re: Scheme (a shorter one)
« Reply #5 on: October 24, 2012, 01:10:58 PM »
Yes it is benzoyl chloride, but why couldn't it be benzoic acid? Because of poor yield, as it is a weak acid?
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Offline discodermolide

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Re: Scheme (a shorter one)
« Reply #6 on: October 24, 2012, 01:17:19 PM »
It is esterifying a phenol which are less reactive than normal alcohols towards estification. Then you need an acid chloride's reactivity to do this reaction.
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Offline Rutherford

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Re: Scheme (a shorter one)
« Reply #7 on: October 24, 2012, 01:25:06 PM »
I will write C to make it easier :rarrow: OH-C6H4-COCH2NH2. Now, how to rationalize the conversion from B to C?
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