I will assume that the sign ? between the C and the CCH3 is a triple bond, and I will represent it by three vertical lines |||
So the reaction is:
<span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->
This is a typical reaction known as halogenation of alkines.
This is an addition reaction, i.e. the alkyne undergoes an addition of the Br2 (and it also happens with Cl2) to the triple bond to form a tetra halide.
.
Br Br
</span> | |
<span><span>CH3CH2CH2CH2C ||| CCH3+2Br2 ----> CH3 CH2 CH2 CH2 C - C</span> - CH3
| |
Br Br
</span>
First you calculate the concentration of [OH⁻] in <span>solution :
POH = - log [ OH</span>⁻]
POH = - log [ 0.027 ]
POH = 1.56
PH + POH = 14
PH + 1.56 = 14
PH = 14 - 1.56
PH = 12.44
hope this helps!
Answer:
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