Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
$(a)$ $C_{6}H_{5}OH$
$(b)$ $C_{6}H_{5}NO_{2}$
$(c)$ $CH_{3}CH=CH-CHO$
$(d)$ $C_{6}H_{5}CHO$
$(e)$ $C_{6}H_{5}CH_{2}^{+}$
$(f)$ $CH_{3}CH=CH-CH_{2}^{+}$

(N/A) The resonance structures are drawn by shifting electrons using curved arrows to show the delocalization of $\pi$ electrons or lone pairs.
$(a)$ Phenol $(C_{6}H_{5}OH)$: The lone pair on oxygen delocalizes into the ring,creating ortho and para negative charges.
$(b)$ Nitrobenzene $(C_{6}H_{5}NO_{2})$: The $\pi$ electrons of the ring delocalize towards the electron-withdrawing nitro group,creating ortho and para positive charges.
$(c)$ $CH_{3}CH=CH-CHO$: The $\pi$ electrons of the $C=C$ bond shift towards the carbonyl group,creating a carbocation at the $C-2$ position.
$(d)$ Benzaldehyde $(C_{6}H_{5}CHO)$: Similar to nitrobenzene,the ring $\pi$ electrons delocalize towards the carbonyl oxygen,creating positive charges at ortho and para positions.
$(e)$ Benzyl carbocation $(C_{6}H_{5}CH_{2}^{+})$: The $\pi$ electrons of the ring delocalize to stabilize the positive charge on the benzylic carbon.
$(f)$ $CH_{3}CH=CH-CH_{2}^{+}$: The $\pi$ electrons of the $C=C$ bond shift to stabilize the positive charge on the terminal carbon.