Preparation of Ethers Questions in English

(C) Williamson's synthesis involves the reaction of an alkyl halide with a sodium alkoxide $(R-X + R'-ONa \rightarrow R-O-R' + NaX)$.
If $R$ and $R'$ are the same,a symmetrical ether is formed.
If they are different,an unsymmetrical ether is formed.
Thus,both types of ethers can be prepared.

(B) The reaction between a $1^o$ alkyl halide and an alkoxide is known as the $Williamson$ synthesis.
$R - X + R' - O^- Na^+ \to R - O - R' + NaX$
This reaction primarily yields an ether as the major product.

(A) The reaction between a primary alkyl halide $(CH_3Br)$ and a strong nucleophile like sodium tert-butoxide $(NaOC(CH_3)_3)$ proceeds via an $S_N2$ mechanism to form an ether.
The reaction is: $CH_3-Br + NaOC(CH_3)_3 \to CH_3-O-C(CH_3)_3 + NaBr$.
This is a classic example of the Williamson ether synthesis.

(C) Williamson's synthesis involves the reaction of an alkyl halide with a sodium alkoxide to form an ether.
For the reaction to proceed via an $S_N2$ mechanism,the alkyl halide should be primary $(1^\circ)$ to minimize steric hindrance.
If a tertiary alkyl halide is used,elimination (forming an alkene) predominates over substitution.
Therefore,to prepare $tert$-butyl methyl ether,we should use a primary alkyl halide $(CH_3Br)$ and a tertiary alkoxide $((CH_3)_3CONa)$.
This corresponds to option $C$.

(C) The synthesis of ethers via the Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide.
To prepare a bulky ether like $(CH_3)_3C-O-CH_3$,it is essential to use a primary alkyl halide $(CH_3-Br)$ and a tertiary alkoxide $((CH_3)_3C-OK)$ to avoid the competing elimination reaction that would occur if a tertiary alkyl halide were used.
Therefore,the reaction $(CH_3)_3C-OK + CH_3-Br \to (CH_3)_3C-O-CH_3 + KBr$ is the most efficient method.

(C) Williamson$'$s synthesis involves an $S_N2$ reaction between an alkoxide ion and an alkyl halide.
Vinyl chloride $(CH_2=CH-Cl)$ is not used in this synthesis because the $C-Cl$ bond possesses partial double bond character due to resonance.
This makes the carbon-halogen bond stronger and shorter,thereby preventing the nucleophilic substitution reaction required for the synthesis.

Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide via an $S_{N}2$ mechanism. To minimize steric hindrance,the alkyl halide should be primary. Therefore,ethanol is converted to bromoethane,and $3-$methylpentan$-2-$ol is converted to its sodium alkoxide.
$1.$ Conversion of ethanol to bromoethane:
$C_{2}H_{5}OH + HBr \to C_{2}H_{5}Br + H_{2}O$
$2.$ Conversion of $3-$methylpentan$-2-$ol to sodium $3-$methylpentan$-2-$oxide:
$CH_{3}CH(CH_{3})CH(OH)CH_{2}CH_{3} + Na \to CH_{3}CH(CH_{3})CH(ONa)CH_{2}CH_{3} + \frac{1}{2}H_{2}$
$3.$ Williamson synthesis reaction:
$CH_{3}CH(CH_{3})CH(ONa)CH_{2}CH_{3} + C_{2}H_{5}Br \to CH_{3}CH(CH_{3})CH(OC_{2}H_{5})CH_{2}CH_{3} + NaBr$
(Product: $2-$ethoxy$-3-$methylpentane)

(N/A) $(i)$ $CH_3CH_2CH_2ONa$ ($Sodium$ $propoxide$) $+$ $CH_3CH_2CH_2Br$ $(1-Bromopropane)$ $\to$ $CH_3CH_2CH_2-O-CH_2CH_2CH_3$ $(1-Propoxypropane)$ $+$ $NaBr$
$(ii)$ $C_6H_5ONa$ ($Sodium$ $phenoxide$) $+$ $C_2H_5Br$ $(Bromoethane)$ $\to$ $C_6H_5-O-C_2H_5$ $(Ethoxybenzene)$ $+$ $NaBr$
$(iii)$ $CH_3-C(CH_3)_2-ONa$ ($Sodium$ $2-methylpropan-2-olate$) $+$ $CH_3Br$ $(Bromomethane)$ $\to$ $CH_3-C(CH_3)_2-OCH_3$ $(2-Methoxy-2-methylpropane)$ $+$ $NaBr$
$(iv)$ $CH_3CH_2ONa$ ($Sodium$ $ethoxide$) $+$ $CH_3Br$ $(Bromomethane)$ $\to$ $CH_3CH_2-O-CH_3$ $(1-Methoxyethane)$ $+$ $NaBr$

(N/A) The formation of ethers by acid-catalyzed dehydration of alcohols is a bimolecular nucleophilic substitution $(S_N2)$ reaction,which involves the attack of an alcohol molecule on a protonated alcohol molecule.
For this reaction to occur,the alkyl group of the alcohol must be unhindered to allow the nucleophilic attack.
In the case of secondary or tertiary alcohols,the alkyl group is sterically hindered.
Due to this steric hindrance,the elimination reaction (forming alkenes) becomes more favorable than the substitution reaction (forming ethers).
Therefore,acid dehydration of secondary or tertiary alcohols primarily yields alkenes instead of ethers.

(N/A) The preparation of ethers by the dehydration of alcohols depends on the reaction conditions,specifically temperature and the nature of the alcohol.
$1.$ Dehydration of $1^{\circ}$ alcohols: Alcohols are heated in the presence of protic acids like $H_2SO_4$ or $H_3PO_4$ to form ethers.
$2.$ Example: Dehydration of ethanol $(CH_3CH_2OH)$:
- At $443 \ K$ in the presence of concentrated $H_2SO_4$,ethanol undergoes $\beta$-elimination to form ethene $(CH_2=CH_2)$ as the major product.
- At $413 \ K$ in the presence of concentrated $H_2SO_4$,ethanol undergoes bimolecular nucleophilic substitution $(S_N2)$ to form ethoxyethane $(C_2H_5-O-C_2H_5)$ as the major product.
Mechanism for ether formation $(S_N2)$:
Step $1$: Protonation of alcohol: $CH_3CH_2OH + H^+ \rightarrow CH_3CH_2OH_2^+$
Step $2$: Nucleophilic attack by another molecule of alcohol: $CH_3CH_2OH + CH_3CH_2OH_2^+ \rightarrow CH_3CH_2-O^+(H)-CH_2CH_3 + H_2O$
Step $3$: Deprotonation: $CH_3CH_2-O^+(H)-CH_2CH_3 \rightarrow CH_3CH_2-O-CH_2CH_3 + H^+$

(N/A) $(i)$ Methoxybenzene: Sodium phenoxide $(C_6H_5ONa)$ + Methyl chloride $(CH_3Cl)$ $\rightarrow$ Methoxybenzene $(C_6H_5OCH_3)$ + $NaCl$
$(ii)$ Methoxyethane: Sodium methoxide $(CH_3ONa)$ + Ethyl chloride $(CH_3CH_2Cl)$ $\rightarrow$ Methoxyethane $(CH_3OCH_2CH_3)$ + $NaCl$
$(iii)$ Ethoxybenzene: Sodium phenoxide $(C_6H_5ONa)$ + Ethyl chloride $(CH_3CH_2Cl)$ $\rightarrow$ Ethoxybenzene $(C_6H_5OCH_2CH_3)$ + $NaCl$
$(iv)$ Ethoxyethane: Sodium ethoxide $(CH_3CH_2ONa)$ + Ethyl chloride $(CH_3CH_2Cl)$ $\rightarrow$ Ethoxyethane $(CH_3CH_2OCH_2CH_3)$ + $NaCl$

(N/A) The Williamson synthesis involves the reaction of an alkyl halide with a strong nucleophile,the alkoxide ion.
Alkoxide ions are strong bases as well as strong nucleophiles.
When a tertiary alkyl halide is used as a substrate,the alkoxide ion acts primarily as a base rather than a nucleophile.
This leads to an elimination reaction (specifically $E2$ mechanism) instead of a substitution reaction.
Consequently,the major product formed is an alkene rather than the desired ether.
For example,the reaction between sodium tert-butoxide and tert-butyl bromide yields $2$-methylpropene as the major product instead of di-tert-butyl ether.

(A) Williamson ether synthesis involves an $S_N2$ reaction between an alkoxide ion and an alkyl halide.
Since $S_N2$ reactions are highly sensitive to steric hindrance,the reaction proceeds most efficiently with primary alkyl halides.
Secondary alkyl halides may undergo elimination reactions,and tertiary alkyl halides almost exclusively undergo elimination to form alkenes instead of ethers.

(D) Ethers can be prepared by Williamson Synthesis,in which an alkyl halide is treated with sodium alkoxide.
The reaction is: $R-X + NaOR \longrightarrow R-O-R + NaX$

(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with a primary alkyl halide to form an ether.
$(P)$ $C_6H_5CH_2Br + CH_3ONa \rightarrow C_6H_5CH_2OCH_3 + NaBr$. This is a $S_N2$ reaction between a primary benzylic halide and a strong nucleophile,yielding an ether as the major product.
$(Q)$ $C_6H_5ONa + CH_3Br \rightarrow C_6H_5OCH_3 + NaBr$. This is a $S_N2$ reaction between a phenoxide ion and a primary alkyl halide,yielding an ether as the major product.
$(R)$ $(CH_3)_3CCl + CH_3ONa \rightarrow (CH_3)_2C=CH_2 + CH_3OH + NaCl$. Since $(CH_3)_3CCl$ is a tertiary alkyl halide,the strong base $CH_3ONa$ promotes an $E2$ elimination reaction,giving an alkene as the major product.
$(S)$ $C_6H_5CH=CHCl + CH_3ONa$. Vinyl halides are unreactive towards $S_N2$ reactions due to the partial double bond character of the $C-X$ bond and the instability of the vinylic carbocation,so no ether is formed.
Therefore,both $(P)$ and $(Q)$ give an ether as the major product.

(A) Williamson's synthesis is a general method for the preparation of symmetrical and unsymmetrical ethers.
In this reaction,an alkyl halide reacts with a sodium alkoxide to form an ether via an $S_N2$ mechanism.
The general reaction is: $R-X + R'-O^- Na^+ \rightarrow R-O-R' + NaX$.
For example,the reaction of an alkyl halide with sodium ethoxide: $R-X + C_2H_5O^- Na^+ \rightarrow R-O-C_2H_5 + NaX$.