Answer:
Sigma Bonds = 14
Pi bonds = 0
Explanation:
Butanol:
[tex]\sf CH_{3} -CH_{2}-CH_{2}-CH_{2}-OH[/tex]
There is no pi bond in this compound (since there is no double or triply bond)
So,
Pi bond = 0
Now, Coming towards sigma bonds
Sigma bonds = 3 (First carbon with hydrogen) + 2 (Second C with H) + 2 (Third C with H) + 2 (Fourth C with H) + 1 (Bond of O and H) + 4 (Bonds between C and C and C and O)
Sigma Bonds = 14
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
