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Saturday, February 1, 2020

What are the groups that LiAlH4 can and cannot reduce?

Lithium aluminium hydride (LiAIH4) is a nucleophilic reducing agent, and used to reduce polar multiple bonds like >C=O, -CN bond because it is hydride (H-) donor. And it is simply abbreviated as LAH.
LIAIH4 can reduce:
(1) Aldehydes to primary alcohols,
(2) Ketones to secondary alcohols,
(3) Carboxylic acids and esters to primary alcohols.
(4) Acid chlorides to primary alcohols.
(5) Ester to primary alcohols.
(6) Acid anhydrides to primary alcohols.
(7) Amides and nitriles to primary amines.
(8) Isonitriles to secondary amines.
(9) Epoxides (Cyclic ethers) to alcohols.
(10) lactones (Cyclic esters) to diols. 

Important Notes:
LiAH4 can reduce Aldehydes, ketones, carboxylic acids, ester, acid chlorides, acids anhydrides, acid amides,  Nitro Compounds and nitriles and isonitriles  without attacking (Isolated non polar bonds , like -C=C- ) double bonds present in the Compounds.
The only exception in this case alpha- Beta unsaturated compounds containing phenyl group in the Beta position, in this case LiAH4 also attack on double bond.
We can say that the double or triple bonds in conjugation with the polar multiple bonds can be reduced.



Oxymercuration demercuration hydration procedure is superior to acid- catalysed hydration of most alkenes. Why ?

Oxymercuration demercuration hydration procedure is superior to acid- catalysed hydration of most alkenes. Because (1) the two stage process of oxymercuration demercuration is fast and takes place under mild condilions and gives more than 90 % yield of alcohol and (2) rearrangement doesn't take place hence most desire product easily obtained.

More reated Questions:

Arrange the following groups in order of their increasing tendency for nucleophilic addition reactions -COO-, -COOH, -COOCH3, -COCH3, -CHO, -COCI, -CONH2. Justify your answer.


The tendency, for nucleophilic additions of a carbonyl group is increased with increasing partial positive charge on the carbonyl carbon atom, which is maximum in -COCl and minimum in -COO-. Thus, the required order is:



      "-COCI > -CHO > -COCH3 > -COOCH3 > -CONH2 > -COOH > -COO-"

More reated Questions:

Friday, January 31, 2020

Prpare Benzilic acid (1-hydroxyl-1,1-diphenylethanoic acid) from either cis or trans stilbene.



First Convert stilbene into vicinal alcohol by oxidation with cold KMnO4 and on acidification gives Benzilic which undergoes rearrangment to form Benzilic acid

More reated Questions:

50 Cm3 of 0.04 M K2Cr2O7 in acidic medium oxidized a sample of H2S gas to Sulphur . The volume of 0.03 M KMnO4 required to Oxidize the same amount of H2S gas to Sulphur, in acidic medium is.


Related Questions:

Calculate the moles of KMnO4 required to reacting with 180 gm of Oxalic acid (H2C2O4). Also calculate the volume of CO2 at STP produce in the reaction? (K=39, Mn=55, Of=16)