Answer: A
At the double bond, C2 carries CHX3 over H and C3 carries C4 over H; the drawing has these higher groups on opposite sides, so the bond is (E). At C4 the priorities are CH=CHX− (C attached to C,C,C with duplication) >CX2HX5 (C,H,H) >CHX3>H; with H away from the viewer the sequence runs clockwise, so C4 is (R). The compound shown is (2E,4R)-4-methylhex-2-ene; it is one of 2×2=4 stereoisomers.