Answer:
b. hydride shift from C-3 to C-2.
Explanation:
Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).
This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.
C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.
it is because it reacts readily with other compounds
The arrow that are drawn correctly are the cool air current and the upper air current. In the sea breeze, warm air moves up because it is less dense that the cool air. The cool air moves down to replace the warm air. The land breeze current moves from the land to the sea while the sea breeze moves from the sea to the land. This creates a continuous flow of current at the sea level.
N2<span> + 3H</span>2<span> = 2NH</span><span>3
so, NH3 = (N2 + 3H2)/ 2
= (28g + 3*25g)/2
= 51.5g</span>
Thank you for the free 15 points .