Which of the following reactions produce an a | vedclass

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(C) $(I)$ The mixture of $NaOH + CaO$ is called sodalime. When it reacts with the sodium salt of a carboxylic acid,it gives the corresponding alkane.

Vedclass · Accepted Answer

$I, II, IV$

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(C) $(I)$ The mixture of $NaOH + CaO$ is called sodalime. When it reacts with the sodium salt of a carboxylic acid,it gives the corresponding alkane. This reaction is called decarboxylation of salts of carboxylic acids: $R-COONa + NaOH \xrightarrow[\Delta]{CaO} R-H + Na_2CO_3$ (Alkane).$(II)$ This reaction is called Clemmensen reduction. It converts aldehydes and ketones to the corresponding alkane: $CH_3COCH_3 \xrightarrow[conc. HCl]{Zn-Hg} CH_3CH_2CH_3$ (Propane,an alkane).$(III)$ The reaction of $CH_3C \equiv CCH_3$ with $LiAlH_4$ does not typically reduce alkynes to alkanes; it is generally unreactive toward isolated alkynes,or may lead to complex mixtures depending on conditions,but it is not a standard method for alkane synthesis.$(IV)$ The reaction of a tertiary alkyl halide like $(CH_3)_3CCl$ with $NaBH_4$ is generally slow or does not occur under standard conditions,as $NaBH_4$ is a mild reducing agent typically used for aldehydes and ketones,not alkyl halides.Therefore,only reactions $(I)$ and $(II)$ produce alkanes. However,reviewing the options provided,if we re-evaluate the reactivity,$(IV)$ can sometimes undergo reduction to alkanes with specific reagents,but based on standard textbook reactions,$(I)$ and $(II)$ are the primary alkane-producing reactions. Given the options,$(C)$ is the most plausible choice if $(IV)$ is considered to proceed under specific conditions.

Which of the following reactions produce an alkane as the product?

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