O-chem destroyer question #86

[navneetdh]

This is question 86 from o-chem destroyer 2008

in the solutions they explained in part a) the conversion of an amide to a primary amine.

CH3--C(O)--NH2 -------> CH3---CH2----NH2

reagents were 1. LiAlH4 and Et2OH and 2. H2O

I thought the final molecule should be

CH3----C(OH)H---NH2 i.e it should have a OH group and a H instead of 2 Hydrogens because LAH only reduces COOH to alcohols and not alkanes. I dont get this. Is that a typo in the book or am I wrong?

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[ZH1988]

This is question 86 from o-chem destroyer 2008

in the solutions they explained in part a) the conversion of an amide to a primary amine.

CH3--C(O)--NH2 -------> CH3---CH2----NH2

reagents were 1. LiAlH4 and Et2OH and 2. H2O

I thought the final molecule should be

CH3----C(OH)H---NH2 i.e it should have a OH group and a H instead of 2 Hydrogens because LAH only reduces COOH to alcohols and not alkanes. I dont get this. Is that a typo in the book or am I wrong?

no the book is right LAH is very strong it will reduce it to an alkane, and plus you dont have a COOH, u only have a C=O

 

[navneetdh]

no the book is right LAH is very strong it will reduce it to an alkane, and plus you dont have a COOH, u only have a C=O

oh right no COOH...my fault i am seeing C=O but thinking COOH lol.

and good to know they will reduce to alkanes. so as a general rule can i take it that if i see COOH, esters, C=O, C-OH and LAH then I can make alkane??

because in a lot of questions I saw COOH only taken to alcohol level even with LAH. Was it diff in this case because of the C=O and not COOH??

so when do we know it will be alcohol and not an alkane?

i also didn't see in any of my study aids about LAh taking the above compounds to alkanes. But I am no genius at O-chem either.

 

[loveoforganic]

Amides are the only case with LiAlH4 where you reduce to alkane. It's a special case. This happens due to aluminum's affinity for oxygen coupled with the stability of the iminium ion.

Here's the mechanism

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-2-4.html

 

[UCB05]

oh right no COOH...my fault i am seeing C=O but thinking COOH lol.

and good to know they will reduce to alkanes. so as a general rule can i take it that if i see COOH, esters, C=O, C-OH and LAH then I can make alkane??

because in a lot of questions I saw COOH only taken to alcohol level even with LAH. Was it diff in this case because of the C=O and not COOH??

so when do we know it will be alcohol and not an alkane?

i also didn't see in any of my study aids about LAh taking the above compounds to alkanes. But I am no genius at O-chem either.

It will never be an alkane. LAH will reduce a carboxy group to a primary alcohol. The only difference is that in this case of an amide, the -OH group is subbed out for a -NH2 group. A primary amine is analagous to a primary alcohol.
(Unless you consider an alcohol an alkane too. then we're just arguing semantics...)

 

[navneetdh]

thanks guys!! makes sense now!!