Question

Match the compounds/reactions in List I with their stereochemistry in List II. Matching can be one or more than one.

Answer 1

Whether an alkyl halide will undergo an $S_{N}1$ or an $S_{N}2$ reaction depends upon a number of factors. Some of the more common factors include the natures of the carbon skeleton, the solvent, the leaving group, and the nature of the nucleophile.
Only those molecules that form extremely stable cations undergo $S_{N}1$ mechanisms. Normally, only compounds that yield $3$ (tertiary) carbonications (or resonancestabilized carbocations) undergo $S_{N}1$ mechanisms rather than $S_{N}2$ mechanisms. Carbocations of tertiary alkyl halides not only exhibit stability due to the inductive effect, but the original molecules exhibit steric hindrance of the rear lobe of the bonding orbital, which inhibits $S_{N}2$ mechanisms from occurring. Primary alkyl halides, which have little inductive stability of their cations and exhibit no steric hindrance of the rear lobe of the bonding orbital, generally undergo $S_{N}2$ mechanisms.
Polar protic solvents such as water favor $S_{N}1$ reactions, which produce both a cation and an anion during reaction. These solvents are capable of stabilizing the charges on the ions formed during solvation. Because $S_{N}2$ reactions occur via a concerted mechanism (a mechanism which takes place in one step, with bonds breaking and forming at the same time) and no ions form, polar protic solvents would have little effect upon them. Solvents with low dielectric constants tend not to stabilize ions and thus favor $S_{N}2$ reactions. Conversely, solvents of high dielectric constants stabilize ions, favoring $S_{N}1$ reactions.
$S_{N}2$ and $E_2$ reactions require a good nucleophile or a strong base. $S_{N}1$ and $E_1$ reactions occur with strong bases with molecules whose -carbon is secondary or tertiary and in the absence of good nucleophiles.
$S_{N}1$ and $E_1$ reactions are not synthetically useful because they almost always give a mixture of substitution and elimination products. The proportion of these products does vary with the -carbon branching, however. Generally speaking, greater branching gives more elimination products. These elimination products generally follow Saytzeff's rule. Keep in mind that the -carbon must be secondary or tertiary for $S_{N}1$ or $E_1$ to occur at all. Thus the branching effect is much less pronounced than in the $S_{N}2$ and $E_2$ reactions.
In the broader context of all substitution and elimination reactions, remember that $S_{N}1$ and $E_1$ will not occur in the presence of a strong base or a good nucleophile. In these cases, $E_2$ and $S_{N}2$ dominate their unimolecular cousins.
$E_{1}cb$ reactions are quite different from $E_1$ reactions. $E_1$ reactions require a strong leaving group and a stable cationic intermediate. The leaving group first leaves to create the cationic intermediate. Then, the elimination occurs.
$E_{1}cb$ is the accepted mechanism for the aldol condensation.
Key Requirements:

• There must be an acidic proton. In the case of the one depicted above, the carbonyl acidifies the alpha-carbon proton enough for it to be deprotonated.
• There usually is a bad leaving group. If it were a really good leaving group, either $E_1$ or $E_2$ would probably occur instead.

The key differences between the two is that $E_1$ goes through a cationic intermediate, while $E_{1}cb$ goes through a anionic intermediate. Therefore, never propose a $E_{1}cb$ in acid or a $E_1$ in base.Also, the $cB$ in $E_{1}cb$ stands for conjugate base meaning the elimination goes through the intermediate that is the conjugate base of my starting material. In this case, the enolate was the conjugate base.