is an extremely useful reagent for organic synthesis in instances where an alcohol needs to be converted to a good leaving group (bromine is an excellent leaving group).  reacts selectively with alcohols, without altering any other common functional groups. This makes it ideal for situations in which a molecule contains acid-sensitive components that prevent the use of a strong acid to protonate a target alcohol.

While the mechanisms differ,  reactions are similar to S_N2 reactions in that they both invert the configuration at the site of attack. The configuration about the carbon adjacent to the alcohol in the given reactant is S. After substitution, the configuration of the major product is R, as is the case in molecule IV.

Based on the given reagents and the specification that the reaction takes place in a single step, it may be concluded that the reaction occurs by an S_N2 or E2 mechanism. Since the compound lacks any moderately acidic hydrogen, an S_N2 reaction is more likely. The absolute configuration at the reaction site in the initial compound is S, which is converted to R as a result of the "back-side attack" characteristic of all S_N2 reactions. The major product is shown below:

This problem involves the synthesis of a Grignard reagent. Grignard reagents are easily created in the presence of halo-alkanes by adding magnesium in an inert solvent (in this case ). Once we have created our Gringard, it can readily attack a carbonyl. In this case, our Grignard attacks carbon dioxide to create our desired product.

An  reaction is most efficiently carried out in a protic solvent. An inverted configuration site is characteristic of an  reaction and the substituted nucleophile does not form a pi bond in an  reaction.

In this question, we're given the reactant and product as well as the reagent being used in the reaction, and we're being asked to identify which reaction mechanism will correctly lead us from reactant to product.

To begin, it's important to notice that the reactant contains a tertiary bromine and the product contains a methoxy group in place of where the bromine was. Thus, we can conclude that a substitution reaction has taken place. If an elimination reaction had taken place, then there would have been a double bond in the product.

Now we need to identify which kind of substitution has occurred. Since the leaving group is attached to a tertiary carbon, we know that a stable carbocation will be generated upon dissociation. Therefore, we would expect this to be an  reaction.

A solvolysis reaction is simply an  reaction where the solvent acts as a nucleophile.

In this case, we start with a tertiary alkyl halide. Bromine, a stellar leaving group, leaves the substrate and leaves a carbocation intermediate. The methanol is then free to attack the carbon chain at the site of the carbocation to form an ether. The correct answer is 2-methoxy-2-methylpropane.

The correct answer is .

Alcohol is a horrible leaving group.  is often employed to convert an alcohol group into a bromine group so that additional substitution and elimination reactions can ensue.

Cyanide is a weak base and a good nucleophile, and the solvent is aprotic; therefore, the  product is favored. This involves 100% inversion of stereochemistry; therefore II is favored.

The double bond in 3-iodo-1-butene would stabilize the transition state through resonance. This would make the transition state lower in energy due to its increased stability. Thus, the reaction would be faster as reactions with lower activation energies proceed at faster rates. 

This reaction would use an  S_N1 mechanism because the leaving group, bromine, is on a tertiary carbon, which is a carbon attached to three other carbon atoms. The bulk of these methyl groups would make S_N2 impossible, but it would make the carbocation produced by an  S_N1 reaction very stable. The methyl group would lead to hyperconjugation, which is a type of resonance that stabilizes transition states.