E1 will not happen here because for E1 to occur, you need a mild to strong base and Br- is a good nucleophile because it has electrons to donate. However, since it is a Halogen and down in the periodic table, it does not have basic character, a paradoxical trend for the nucleophilic halogens (as you'd think that more basic, more nucleophilic). The answer would be E because the carbon with the chlorine leaving group is tertiary and will form a carbocation. Because it forms a carbocation it has sp2 configuration and therefore can be attacked from either side. Also, the tertiary carbon is chiral. Thus, enantiomers will form, yielding a racemic mixture, and loss of optical activity. Hope that helps.