The elimination reaction (as opposed to the SN2 because of the big bulky base reagent) only proceeds with the anti-periplanar product. 

Because both hydrogens neighboring the  are anti-periplanar, both elimination products would be formed. 

For E2 reactions a tertiary electrophile > secondary electrophile > primary electrophile. A polar aprotic solvent favors E2 (remember that polar protic solvents favor E1). A strong base is needed to undergo E2.

SN2 reactions result in an inversion of products, thus R-configured reactants become S-configured products and vice versa. Polar aprotic solvents such as ether favor SN2 reactions.

E1 and E2 are elimination mechanisms and the question asks for a substitution mechanism. SN1 results in racemization of products, not inversion. Thus, if the reaction had gone through SN1 we would see a mixture of R and S-configuration in the products. Additionally, polar protic solvents (not aprotic like ether) favor SN1 reactions.

Here we have a classic electrophilic addition reaction.

We must remember that a double bond is an area in a molecule that has a high electron density. As a result, double bonds can act as nucleophiles and react with good electrophiles.

In this problem, the alkene attacks the dibromide to form a bromonium ion, which is rather unstable. The bromonium ion pulls carbon's electrons toward itself, causing the carbons in the bromonium complex to have a slightly negative charge. As a result, the negatively charged bromine can attack either of the carbon atoms and, thus, we have added two bromines to our carbon chain.

The five resonance structures of the given molecule are shown below:

The nitro substituent is strongly deactivating, ketones are moderately deactivating substituent, halides are weakly deactivating, alkyl groups are weakly activating, and amine groups are strongly activating.

An SN1 mechanism involves the leaving of the bromine in the first step and the formation of a carbocation on that carbon. The molecule with the most stable carbocation will react most quickly.

The carbocation is most stable on the most highly substituted carbon. All the of the answer options form primary or secondary cations except for one. The correct answer has a carbocation on a carbon bonded to two methyl groups and one ethyl group. The correct answer is .

Grignard reagents attack ketones at the site of the ketone's carbon to produce tertiary alcohols.

SN2 reactions require a strong nucleophile. If the the nucleophile is weak, SN1 is favored over SN2.

For SN2 reactions less substitution in the electrophile is favored (methyl > primary > secondary).

For SN2 reactions polar aprotic solvents are preferred (DMSO, acetone, DMF and ethers are common polar aprotic solvents). Polar protic solvents (ethanol, methanol) favor SN1 reactions.