structure. It, too, has 26 valence electrons, and again only chlorine has unshared CHEMICAL BONDING (d) This compound has the same molecular formula as the compound in part (c), but a different H (e) The constitution of CHaNHCH, CH3 is shown(below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair(below right) HH (Oxygen has two unshared pairs in(CH3),CHCH=O. H H H C=0 1. 13 (b) This compound has a four-carbon chain to which are appended two other carbons. is equivalent to CH- which may be rewritten as (CH,)2CHCH( CH (c) The carbon skeleton is the same as that of the compound in part(b), but one of the terminal carbons bears an OH group in place of one of its hydrogens HO—C-H which may be CHOH is equivalent to written as 一C CH CHCH(CH3) (d) The compound is a six-membered ring that bears a-C(CH3)3 substituent. HH H-C-C H is equivalent to which may be CH rewritten as C(CH3)3 CH3 HH CH3 problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and of the carbon-oxygen bonds is a double bond. Since a neutral carbon is associated with four Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE WebsiteCHEMICAL BONDING 5 (d) This compound has the same molecular formula as the compound in part (c), but a different structure. It, too, has 26 valence electrons, and again only chlorine has unshared pairs. (e) The constitution of CH3NHCH2CH3 is shown (below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair (below right). ( f ) Oxygen has two unshared pairs in (CH3)2CHCH?O. 1.13 (b) This compound has a four-carbon chain to which are appended two other carbons. (c) The carbon skeleton is the same as that of the compound in part (b), but one of the terminal carbons bears an OH group in place of one of its hydrogens. (d) The compound is a six-membered ring that bears a @C(CH3)3 substituent. 1.14 The problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and one of the carbon–oxygen bonds is a double bond. Since a neutral carbon is associated with four which may be rewritten as H C C C C C C H H C H H H H H H H H CH3 CH3 CH3 is equivalent to C(CH3)3 CH2OH CH3CHCH(CH3)2 H H CH3 CH3 C HO H C CH3 H C is equivalent to which may be rewritten as HO CH3 CH3 H H CH3 C C CH3 (CH3)2CHCH(CH3)2 is equivalent to which may be rewritten as H H H C H H C H H C H C O H H H C H H H C H H H N C H H H C H H H C H H N C H H H C H H C Cl Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website