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We need to identify which method does NOT correctly prepare alcohols.
Option 1:
This is a correct method. Grignard reagents ($$RMgBr$$) are powerful nucleophiles that attack the carbonyl group of ketones. The mechanism involves nucleophilic addition of $$R^-$$ to the electrophilic carbonyl carbon, forming an alkoxide intermediate. Subsequent hydrolysis (treatment with dilute acid) gives a tertiary alcohol:
$$R'_2C=O + RMgBr \rightarrow R'_2(R)C-OMgBr \xrightarrow{H_3O^+} R'_2(R)C-OH$$This is one of the most important methods for synthesizing alcohols.
Option 2:
This is a correct method. The hydroxide ion ($$OH^-$$) acts as a nucleophile and attacks the carbon bearing the halogen in an $$S_N2$$ (or $$S_N1$$) reaction, displacing the halide ion:
$$R-X + NaOH(aq) \rightarrow R-OH + NaX$$This nucleophilic substitution reaction produces an alcohol.
Option 3:
This is a correct method. In this two-step process, the alkene first reacts with borane ($$BH_3$$) to form a trialkylborane (hydroboration), which is then oxidized with alkaline hydrogen peroxide ($$H_2O_2/NaOH$$) to give an alcohol. The reaction gives anti-Markovnikov addition and syn-addition of water across the double bond:
$$R-CH=CH_2 \xrightarrow{BH_3/THF} (RCH_2CH_2)_3B \xrightarrow{H_2O_2/NaOH} RCH_2CH_2OH$$Option 4:
This is an INCORRECT method for preparing alcohols. Ozonolysis involves treating an alkene with ozone ($$O_3$$) followed by a reductive workup (using $$Zn/H_2O$$ or $$(CH_3)_2S$$). This reaction cleaves the carbon-carbon double bond and produces aldehydes and/or ketones, NOT alcohols:
$$R_2C=CR_2 \xrightarrow{O_3} \text{ozonide} \xrightarrow{Zn/H_2O} R_2C=O + O=CR_2$$The products are carbonyl compounds (aldehydes or ketones depending on the substitution), not hydroxyl compounds.
The correct answer is Option 4: Ozonolysis of alkene is an incorrect method for preparing alcohols.
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