Printing og meta only if posts page?

You have nested <?php ?> blocks, that is illegal syntax. To cause HTML output without echo or print statements, you must close out the PHP block, then resume it again.

<?php 
if(is_single()){ ?>
<meta property="og:title" content="<?php the_title_attribute(); ?>" />

Alternately, use echo statements to cause output and never close out and resume PHP.

<?php 
if(is_single()){ 
echo '<meta property="og:title" content="' . the_title_attribute(['echo'=>false,]) . '" />';

Note how the_title_attribute() is supplied with an argument to return values instead of echoing. This amounts to the same thing:

<?php 
if(is_single()){ 
echo '<meta property="og:title" content="';
the_title_attribute();
echo '" />';

PHP will expand variables within double quoted strings, so this works too:

<?php 
if(is_single()){ 
$attr = the_title_attribute(['echo'=>false,]);
echo "<meta property='og:title' content='$attr' />";

If you don’t like single quotes in HTML, replace every occurrence of ' with \"

in other alternative. you just close your php code first. like @bcworkz first answer. in all you need. just replace your code like this

<?php  if(is_single()){ ?>
<meta property="og:title" content="<?php the_title_attribute(); ?>" />
<meta property="og:type" content="article" />
<meta property="og:url" content="<?php echo the_permalink(); ?>" />
<meta property="og:image" content="<?php echo $img_src; ?>" />
<?php };?>